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(B) The intensity at any point in the interference pattern is given by $I = 4I_0 \cos^2(\frac{\phi}{2})$,where $I_0$ is the intensity of each slit and

Vedclass · Accepted Answer

$(2n + 1)\frac{\lambda}{4}$

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(B) The intensity at any point in the interference pattern is given by $I = 4I_0 \cos^2(\frac{\phi}{2})$,where $I_0$ is the intensity of each slit and $4I_0$ is the peak intensity $(I_{max})$.Given that the intensity at a point is half the peak intensity,we have $I = \frac{1}{2} I_{max} = \frac{1}{2} (4I_0) = 2I_0$.Substituting this into the intensity formula: $2I_0 = 4I_0 \cos^2(\frac{\phi}{2})$.This simplifies to $\cos^2(\frac{\phi}{2}) = \frac{1}{2}$,which means $\cos(\frac{\phi}{2}) = \pm \frac{1}{\sqrt{2}}$.Thus,$\frac{\phi}{2} = (2n + 1)\frac{\pi}{4}$,where $n$ is an integer.Since the phase difference $\phi = \frac{2\pi}{\lambda} \Delta x$,we substitute this into the equation:$\frac{1}{2} \cdot \frac{2\pi}{\lambda} \cdot \Delta x = (2n + 1)\frac{\pi}{4}$.Solving for the path difference $\Delta x$,we get $\Delta x = (2n + 1)\frac{\lambda}{4}$.

In the Young's double slit experiment using a monochromatic light of wavelength $\lambda$,the path difference (in terms of an integer $n$) corresponding to any point having half the peak intensity is :-

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